Question

sin^4x+cos^4x=cos^2*2x
${sin}^{4} x + {cos}^{4} x = {cos}^{2} 2x$
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Answer 1

First, let's rewrite the equation using the identity cos^2x = 1 - sin^2x:

sin^4x + (1 - sin^2x)^2 = cos^2*2x

Expanding (1 - sin^2x)^2:

sin^4x + 1 - 2sin^2x + sin^4x = cos^2*2x

Combine like terms:

2sin^4x - 2sin^2x + 1 = cos^2*2x

Next, let's use the double-angle identity for cosine:

cos^2*2x = (cos2x)^2 = (2cos^2x - 1)^2

Substituting this back into the equation:

2sin^4x - 2sin^2x + 1 = (2cos^2x - 1)^2

Expanding (2cos^2x - 1)^2:

2sin^4x - 2sin^2x + 1 = 4cos^4x - 4cos^2x + 1

Rearranging the terms:

2sin^4x - 4cos^4x + 2sin^2x - 4cos^2x + 1 - 1 = 0

2sin^4x - 4cos^4x + 2sin^2x - 4cos^2x = 0

Divide all terms by 2:

sin^4x - 2cos^4x + sin^2x - 2cos^2x = 0

Factor out common terms:

(sin^2x - 2cos^2x) * (sin^2x + 1 - 2cos^2x) = 0

sin^2x - 2cos^2x = 0 or sin^2x + 1 - 2cos^2x = 0

Using the identity sin^2x = 1 - cos^2x:

1 - cos^2x - 2cos^2x = 0

-3cos^2x = -1

cos^2x = 1/3

Taking the square root of both sides:

cosx = ±√(1/3)

sin^2x + 1 - 2cos^2x = 0

sin^2x + 1 - 2(1/3) = 0

sin^2x - 2/3 = 0

sin^2x = 2/3

Taking the square root of both sides:

sinx = ±√(2/3)

So the solutions for x are:

x = arcsin(√(2/3)), x = π - arcsin(√(2/3)), x = π + arcsin(√(2/3)), x = 2π - arcsin(√(2/3))